Trigonometric functions and their properties

Use \((\sin x-\cos x)^2=\sin^2x+\cos^2x-2\sin x\cos x=1-2\sin x\cos x\). Since \(\sin x-\cos x=\frac{1}{2}\), we get \(\frac{1}{4}=1-2\sin x\cos x\). Hence \(2\sin x\cos x=\frac{3}{4}\), so \(\sin x\cos x=\frac{3}{8}\). The value \(\frac{1}{4}\) is the square of the given difference, not the product. Exam tip: square the given expression first and then use \(\sin^2x+\cos^2x=1\).

Use the identity ((\sin x+\cos x)^2+(\sin x-\cos x)^2=2). Hence the value is (2-\frac{36}{25}=\frac{14}{25}).

Given \(\tan x+\cot x=5\). Squaring both sides gives \((\tan x+\cot x)^2=\tan^2x+\cot^2x+2\tan x\cot x\). Since \(\tan x\cot x=1\), we get \(25=\tan^2x+\cot^2x+2\). Hence, \(\tan^2x+\cot^2x=23\). Option 25 is only the square of the given sum; the extra 2 must be subtracted. Exam tip: always use \(\tan x\cot x=1\) in such identities.

Using the identity \(\sec^2 x-\tan^2 x=1\), we get \((\sec x-\tan x)(\sec x+\tan x)=1\). Hence \(\frac{1}{4}(\sec x+\tan x)=1\), so \(\sec x+\tan x=4\). The option \(16\) may result from incorrectly squaring \(\frac{1}{4}\). Exam tip: rewrite \(\sec^2 x-\tan^2 x=1\) as a product in such questions.

The identity is ((\cosec x+\cot x)(\cosec x-\cot x)=1). Therefore, the required value is (\frac{1}{6}).

By the half-angle identity, (\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}). In such forms, identify the denominator carefully.

The standard half-angle form is (\cot \frac{x}{2}=\frac{1+\cos x}{\sin x}). Keep the forms of (\tan \frac{x}{2}) and (\cot \frac{x}{2}) separate.

The range of \(\cos x\) is \([-1,1]\). Adding 2 to both endpoints gives the range \([1,3]\), so the minimum is 1. In exams, when a constant is added, shift both endpoints of the range by that constant.

Since \(\cos x\) lies between \(-1\) and \(1\), the minimum value of \(5\cos x\) is \(-5\). Hence, the minimum value of \(2+5\cos x\) is \(2-5=-3\), attained when \(\cos x=-1\). Option \(7\) is the maximum value, obtained when \(\cos x=1\). Exam tip: for \(a+b\cos x\) with \(b>0\), the minimum value is \(a-b\).

Since \(\sin 2x\) always lies between \(-1\) and \(1\), the range of \(3\sin 2x\) is \([-3,3]\). Adding \(-1\) shifts every value down by 1, giving \([-3-1,\,3-1]=[-4,2]\). Hence, option B is correct. Option A accounts only for multiplication by 3 and misses the downward shift by 1. Exam tip: the range of \(a\sin(bx)+c\) is \([c-|a|,\,c+|a|]\).

For a function of the form \(a\cos(bx)+c\), the period is \(\frac{2\pi}{|b|}\). Here, \(b=3\), so the period is \(\frac{2\pi}{3}\). The coefficient 2 changes only the amplitude, and +4 shifts the graph upward; neither changes the period. \(\frac{\pi}{3}\) is half of the period, so it is not correct. Exam tip: For \(\sin bx\) or \(\cos bx\), divide \(2\pi\) by \(|b|\) to find the period.

The period of (\tan kx) is (\frac{\pi}{k}). Here (k=2), so the fundamental period is (\frac{\pi}{2}).

In the second quadrant, (\sin x) is positive and (\cos x) is negative. Since (\sin x=\frac{4}{5}), (\tan x=-\frac{4}{3}).

In the third quadrant, (\cos x) is negative. Since (\cos x=-\frac{12}{13}), (\sec x=-\frac{13}{12}).

In the fourth quadrant, (\cos x) is positive and (\sin x) is negative. From the (7,24,25) triple, (\cos x=\frac{7}{25}).

(\cos x=-\frac{8}{17}). In the second quadrant, (\sin x) is positive, so (\sin x=\frac{15}{17}).

(2\pi-x) is related to the fourth quadrant. There (\sin x) is negative, so (\sin(2\pi-x)=-\sin x).

(2\pi-x) lies in the fourth quadrant and (\cos x) remains positive. Hence (\cos(2\pi-x)=\cos x).

The angles \(2\pi-x\) and \(-x\) are coterminal because they differ by \(2\pi\). Hence, \(\tan(2\pi-x)=\tan(-x)\). Since tangent is an odd function, \(\tan(-x)=-\tan x\). The \(\cot x\) option is incorrect because tangent changes to cotangent for complementary angles, not for this angle. Exam tip: First reduce \(2\pi-x\) to \(-x\).

In the form (\frac{3\pi}{2}+x), (\sin) changes to (\cos) with a negative sign. Hence the answer is (-\cos x).